Logarithmic endpoint asymptotic of the complete elliptic integral ($K(m)=L+(1-m)(L-1)/4+O((1-m{)}^{2}L),L=\log [4/\sqrt{1-m}]$)

Here $m$ is the parameter, equal to the squared modulus. Near $m=1$ the endpoint integrand is locally $(1-m+s^2{)}^{-1/2}$. Matching its integral to the outer secant integral gives $L$. The next correction is of order $(1-m)\log (1/(1-m))$. Keeping the parameter/modulus convention explicit prevents a factor-of-two error in the complementary small quantity.